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History of Electrical Forces

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									History of Electrical Forces Electron - Greek word meaning amber - Amber is a brownish, semi-transparent sap which flows from softwood trees - When rubbed has "electrical" properties Theories Explaining Electrical Charge One Fluid Model (Benjamin Franklin) - electrical fluid is contained in all substances - excess of fluid means positive charge - deficit means negative charge Two Fluid Theory - two kinds of electrical fluid (positive and negative) - excess of positive fluid creates a positive charge Modern Theory - there are only two types of electrical charge Positive and Negative (proton) (electron) stationary moves - like charges repel; unlike charges attract
1

Electrostatic Conduction - a charged body transfers a charge onto a neutral body - electroscope is left with a net negative charge

Electrostatic Induction - a charged body induces a shift of charge on a nearby neutral body ------Plastic rod
- - - - - - - - - + + + + + + + + + Rubbed with cat fur

- momentary grounding while the negative rod is held nearby - leaves electrosope with net positive charge
2

Lodestone = magnetic iron ore = magnet = compass

- WHY do magnets point to the geographic poles? Ans. Stars, effluvium - The geographic North seeking pole of the magnet (or compass) geographic North seeking pole North seeking pole North pole N pole N  At the geographic North Pole of the earth, there must be (by definition) a magnetic South pole  At the geographic South Pole of the earth, there must be (by definition) a magnetic North pole

3

Gilbert (1600) WHY do magnets point to the geographic poles?

giant lodestone "terella" or little earth

" the earth itself is a GIANT LODESTONE"

4

Magnetic Domains "each atom is a magnet" Which atoms only? IRON NICKEL COBALT

How do we make the domains align in one direction? 1. Let the earth's magnetic field do the job over millions of years. 2. Stroke the unmagnetized magnet with the magnetized magnet. 3. Electromagnet
5

Earth

Sun

Solar Wind

Molten Iron

Nickel

*

- Alpha Particles - protons - electrons - neutrons - x-rays - U.V -absorbed by ozone * Can cause mutations

sunspots

Magnetic Field (can deflect charged particles)

6

Sun

Solar Wind

Earth

Solar Prominences

Sun Spots

7

GRAVITY Why does something fall? 1. Animism - Spirits
2. Aristotle - natural place Fire Air Water Earth

Natural Motion is an object moving to its natural place 3. Newton - Gravity Masses attract each other 4. Einstein - warp is space 5. Super strings 6. ? Future paradigms

8

The space shuttle is traveling in a circular orbit… Thus there is a centrifugal force on the astronaut!

Force up (centrifugal) Astronaut Force down (gravitational) Centripetal If the astronaut is in ORBIT, then the centrifugal force up equals the gravitational force (centripetal) down and he feels apparently weightless because there is no net force on him.
9

Newton's Gravitational Force Law Example Calculate the gravitational force between a 70 kg man and the earth.

Fg 

Gm1m 2 R2

2   11 Nm   6.67 x 10 70 kg  5.98 x 10 24 kg  kg 2    Fg  2 6 6.37 x 10 m average radius due to equatorial bulge Fg = 6.9 x 102 N *this is his weight









OR Fg = mg

 m F g  70 kg  9.81  s2 
Fg = 6.9 x 102N
10

   

Newton's Gravitational Force Law
Fg  Gm1 m 2 R2

If the moon's mass is 1/6 that of the earth, then on the moon, your mass and weight would be…? EARTH
R = 6.37 x 106 m  4000 miles R Average Radius  the length of Canada

R How much would you actually weigh at points A, B, and C? But how much does a space shuttle astronaut weigh only 180 km up?

+A

R +B (2R)

R

+C(3R)
11

Problem #1 Calculate the centrifugal force on a 100 kg man at the earth's surface (at the equator)

Fcent 
Fcent 

4 2 mR T2
2

4 2 ( 100 kg )6.37 x 10 6 m s   24 hr  3600   hr  
= 3.37 N

12

Problem #2 Calculate the revolution period of the moon. (in days) *The moon is 30.2 earth diameters away.

Fg

Fcent

Fup = Fdown Fcent = Fg

4 2 mR T2
2



Gm1m2 R2

4 2 R 3 T  Gm
= 27.5 days *moon is 1.5" farther/year
13

Problem #3 What would the period (in hours) of the earth's rotation have to be in order for a person to feel weightless? (on the equator) Fup = Fdown Fcent = Fg

4 2 mR T2
2



Gm1m2 R2
2 3

4 R T  Gm

T = 1.40 hours

14

Problem #4 An artificial satellite of mass m is moving with constant speed v around the earth in a circular orbit of radius r. What is the speed of the satellite if it is 6.37 x 106 m up? Fup = Fdown Fcent = Fg
m1v 2 Gm1m2  R R2



v

Gm R

6.67 x 10 v

 11

Nm 2 kg 2

x 5.98 x 10 24 kg

2( 6.37 x 10 6 m )
km s km v  2.0 x10 4 hr v  5.60

15

Problem #5 Calculate the radial velocity of the earth at the equator. (in miles per hour)

 1km  2 6.37 x 10 6 m   3   d 2R  10 m  s   t t 24 hr





 1667.6621

km  1mile    hr  1.6 km 
3

mi  1.04 x10 hr

16

An interesting consequence of the inverse square law for gravitational forces is that an astronaut would fee weightless inside a hollow planet. Newton
Fg  Gm1m 2 R2

A

B Hollow Planet

Newton proved mathematically that there was no net force on the astronaut.
17

Ben Franklin (1775) Pith Ball

Charged hollow metal sphere Hanging straight down (pith balls repel) (no net force) Priestly - (studied Newton's work in England) Suggested maybe electrical forces also vary inversely as the square of the distance between them. (reasoned by analogy with the hollow planet example)

18

Charles Augustin Coulomb (1738 - 1806) Relative Charge

Pith Ball A B C D

Assume A starts with a charge Q B, C, D grounded

A A  B Q/2 B  C Q/2 C  D Q/2 D  A 5/16 Q

B Q/2 Q/4 Q/4 Q/4

C 0 Q/4 Q/8 Q/8

D 0 0 Q/8 5/16 Q

Total Q Q Q Q

* Assume that when two objects touch, the charge equally distributes itself! The Law of Conservation of Charge

19

Coulomb's Torsion Balance

A

B

If A was moved 3X closer to B B would swing 9X farther If A was moved 2X farther from B
1 B would swing 4 as far

This proved that electrical forces vary inversely as the square of the distance.
20

- If the charge on pithball A was doubled and the charge on B was tripled, then B would swing 6X farther. - If the charge on A was quadrupled and the charge on B was halved then B would swing 2X farther.  This proved that electrical forces vary directly as the product of the charges.

Fel 

KQ1Q2 R
2

Coulomb's Electrical Force Law
21

Electric Force Law (Coulomb) Example Calculate the electric force between a 2.0 C charge and a 1.6 C charge placed 10.0 m apart.

Fel 

KQ1Q2 R
2

 Nm 2   8.99 x 10 9 2.0C 1.6 C   C2   Fel   10.0 m 2

= 2.9 x 10 N
22

8

Example Calculate the net electric force on B. -3.0 C A 0.20 m +4.2 C B 0.30 m -3.4 C C

Fel 

KQ1Q2 R2

Fel 

KQ1Q2 R2

2   8.99 x 10 9 Nm 3.0 C 4.2C   C2   Fel   0.2 m 2

2   8.99 x 10 9 Nm 4.2C 3.4 C   C2   Fel   0.30 m 2

= 2.832 x 1012 N left

= 1.426 x 1012 N right

Subtract to find net force Answer 1.4 x 1012 N left

23

Example Find the net electrical force on B. -2.0 x 10-6 C A 3.2 m -3.8 C B 4.1 m C -1.5 C

Fel 

KQ1Q2 R

2

2   8.99 x 10 9 Nm  2 x 10 6 C 3.8 x 10 6 C  C2   Fel   3.2 m 2





Fel  

R2 2   8.99 x 10 9 Nm  3.8 x 10 6 C 1.5 x 10 6 C  2  
C 

Fel 

KQ1Q2

4.1m 2

= 6.67 x 10-3 N (down on B)
R A B
R
2 2

= 3.05 x 10-3 N (left on B)
O 6.67 x 10 3 N tan   A 3.05 x 10  3 N

6.67 x10  3 N 2  3.05 x10  3 N 2

= 65

= 7.3 x 10-3 N

3.05 x 10-3 N  180 + 65 = 245 or W 65 S or S 25 W

6.67 x 10-3 N

24

Example Calculate the gravitational force between an electron and a proton placed 1.0 cm apart.
2   11 Nm   6.67 x 10 9.11 x 10  31 kg 1.67 x 10  27 kg kg 2  Gm1m2    Fg   2 R2 1.0 x 10  2 m











= 1.0 x 10-63 N Example Calculate the electrical force between an electron and a proton placed 1.0 cm apart.
2  9 Nm   8.99 x 10 1.6 x 10  19 C 1.6 x 10  19 C C2  KQ1Q2    Fel   2 2 R2 1.0 x 10 m











= 2.3 x 10-24 N

2.3 x 10 24 N 1.0 x 10  63
25

 10 39 X

Direction of a Gravitational Field Definition: Release a mass and see which way it goes.

26

The Direction of an Electric Field - release a positive charge and see which way it goes
+ -

- - - - - -

- + B

A battery pumps charge

+++++++

electrons are pumped from the negative terminal to the positive

Q. What is the direction of the electric field? A + + B + + + + + + C+ +

A - B

C -

27

The Direction of a Magnetic Field - release a North pole and see which way it points N

S

S A B S N (Left of the page) N (up the page)

28

These arrows represent the direction of the magnetic field at various spots around the earth.

29

The Magnitude of a Gravitational Field
Fg  Gm1m 2

R2  Gm1  Fg  m   2    R 
 Gm1 g R2

gravitational field strength
Calculate
 g

for the earth

 Nm 2   6.67 x 10  11  5.98 x 10 24 kg kg 2     g 6 2 6.37 x 10 m









m  g  9.83 2 s
30

Units of g



 F g  mg
 F g  m
N  kg
N m or kg s 2 m

 F g m

s2 N kg m

kg s 2

Example What is the gravitational field strength at a point in space where a proton experiences a 2.4 N force?

2.4 N m  F 27 N g   1.4 x 10 or  27 m 1.67 x 10 kg s 2 kg

31

The Magnitude of an Electric Field

Fel 

KQ1Q2 R2

 F E Q

 KQ1  Fel  Q  2    R 

 KQ E R2

 Fel  QE

Electric Field Strength

Example Calculate the Electric Field Strength 3.2 m from a 6.4 C charge.

 KQ E R2

 Nm 2   8.99 x 10 9 6.4C    C2   E 3.2 m 2  9 N E  5.6 x 10 C
32

Example Calculate the electric field strength midway between a proton and an alpha particle placed 1.0 cm apart. P+ A2+

 KQ E R2
2   8.99 x 10 9 Nm  1.6 x 10  19 C   C2    E 2 5.0 x 10  3 m





 KQ E R2
2   8.99 x 10 9 Nm  1.6 x 10  19 C   C2    E 2 5.0 x 10  3 m





 N E  5.75 x 10  5 C
RIGHT





 N E  1.15 x 10  4 C
LEFT Subtract!





Left Of the Page  towards the proton or away from the alpha particle
33

 5 N E  5.8 x 10 C

Example What is the initial acceleration on a proton placed in an electric field of strength 7.4 x 105 N/C?

 F E Q

 F = F

 F  Eq

 Eq  mA
 Eq A m
 5 N  19 C  7.4 x 10  1.60 x 10 C  1.67 x 10  27 kg









13 m  7.1 x 10 2

s

34

The Magnitude of a Magnetic Field - the unit is named after the Russian Nikoli Tesla - the unit of magnetic field strength is called the Tesla

Newton Tesla  Amp  metre

 Symbol = B
Unit = T
35

Results of Millikan's Experiments 1. 4.8 x 10-19 C 2. 1.6 x 10-19 C 3. 3.2 x 10-19 C 4. 6.4 x 10-19 C 5. 1.6 x 10-19 C 6. 8.0 x 10-19 C 7. 1.6 x 10-19 C 8. 4.8 x 10-19 C 9. 9.6 x 10-19 C 10. 3.2 x 10-19 C 11. 1.6 x 10-19 C 12. 1.6 x 10-19 C 13. 3.2 x 10-19 C "Charge is QUANTIZED"
36

Millikan's Oil Drop Experiment 1909

Fup =Fdown Fel = Fg

  Eq  mg

 v  q  volume x density  g  Anet  d 
4 3 R 3

This is what Millikan was looking for

37


								
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